Let's rewrite and factorise the inequality
#2x^3+x^2>6x#
#2x^3+x^2-6x>0#
#x(2x^2+x-6)>0#
#x(2x-3)(x+2)>0#
Let #f(x)=x(2x-3)(x+2)#
We can build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaa)##0##color(white)(aaaaa)##3/2##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##x+2##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##2x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaaa)##+#
Therefore,
#f(x)>0# when #-2 < x < 0# and #x > 3/2#
Or in interval notation, #x in ]-2,0[uu]3/2,+oo[#